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ASHA 777 [7]
3 years ago
15

light incident on a polarizer is then passed through a second polarizer. if the polarizer and the analyzer are perpendicular to

each other, what can you say about the light waves emerging from the second polarizer?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
Physics
2 answers:
Lerok [7]3 years ago
8 0
<span>Indeed, this is one of the odd results in physics. A system of two polarizing filters arranged as shown below trasmits no light.</span>
rewona [7]3 years ago
8 0

Answer:

The intensity of light emerging from the second polarizer is zero.

Explanation:

The angle between the axis of first polarizer and the second polarizer is 90°.

According to the law of Malus

I = I_{0}\times Cos^{2}\Theta

where, I_{0} be the intensity of incident light on first polarizer and I be the intensity of light emerging out from second polarizer.

The intensity of light emerging out from first polarizer is given by

I_{1}=\frac{I_{0}}{2}

the inetnsity of light emerging out from second polarizer is given by

I = I_{1}\times Cos^{2}90

I = 0

Thus, no light waves are emerging out from the second polarizer.

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You throw a rock straight up and find that it returns to your hand 3.40 s after it left your hand. Neglect air resistance. What
Soloha48 [4]

Answer:

The maximum height of the rock is 14.2 m

Explanation:

The equations that describe the height and velocity of the rock are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

y0 = initial height

t = time

g = acceleration due to gravity (-9.8 m/s² if upward is positive)

v = velocity of the object at time t

We know that at t = 3.40 s, the rock is in your hand again. Then, if we place the origin of the frame of reference at your hand, the position of the rock at 3.40 s is 0 m. Using the equation of the position, we can calculate the initial velocity that we will need to obtain the max-height.

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · 3.40 s - 1/2 · 9.8 m/s² · (3.40 s)²

(1/2 · 9.8 m/s² · (3.40 s)² ) / 3.40 s = v0

v0 = 16.7 m/s

At max-height, the velocity of the rock is 0. Then, using the equation of velocity we can calculate the time it takes the rock to reach the max-height. With that time, we can calculate the maximum height.

v = v0 + g · t      (at max-height, v = 0)

0 = 16.7 m/s - 9.8 m/s² · t

- 16.7 m/s /  - 9.8 m/s² = t

t = 1.70 s

Now, using this time in the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 16.7 m/s · 1.70 s - 1/2 · 9.8 m/s² · (1.70 s)²

y = 14.2 m

The maximum height of the rock is 14.2 m

8 0
3 years ago
Write<br>any<br>to increase<br>magnitude of current in dynamo​
Svetach [21]

Answer:

Explanation:

It can be increased by: increasing the rate of rotation. Increasing the strength of the magnetic field. Increasing the number of turns on the coil.

Hope this helps

plz mark it as brainliest!!!!!!

4 0
3 years ago
A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met
galben [10]
3 meters per second
5 0
3 years ago
Why are household wall sockets alternating current (AC) instead of direct current (DC)? Select all that apply.
Bond [772]

Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

8 0
2 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
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